5 Answers, 1 is accepted
0

Shinu
Top achievements
Rank 2
answered on 09 Oct 2012, 11:55 AM
Hi,
One suggestion is that you can use FormTemplate to open custom for. Here is the demo which implements the same.
aspx:
Thanks,
Shinu.
One suggestion is that you can use FormTemplate to open custom for. Here is the demo which implements the same.
aspx:
<
EditFormSettings
EditFormType
=
"Template"
>
<
FormTemplate
>
. . .
</
FormTemplate
>
</
EditFormSettings
>
Thanks,
Shinu.
0

Robert
Top achievements
Rank 1
answered on 09 Oct 2012, 12:15 PM
Thank you, but what im looking for is not a form inside the grid. I got an insertform on a different page id like to link to. So can i simple cause the "Add new record" click to send me to this page and not open any form, custom or otherwise, in the grid.
I could just add a link outside the grid to this page, but i like how it looks.
Also id like to change the display text for Add new record, if you know how i can do this, id really appreciate it.
Thanks
I could just add a link outside the grid to this page, but i like how it looks.
Also id like to change the display text for Add new record, if you know how i can do this, id really appreciate it.
Thanks
0
Accepted

Jayesh Goyani
Top achievements
Rank 2
answered on 09 Oct 2012, 12:25 PM
Hello,
Thanks,
Jayesh Goyani
protected
void
RadGrid1_ItemCommand(
object
sender, GridCommandEventArgs e)
{
if
(e.CommandName == RadGrid.InitInsertCommandName)
{
// Your logic comes here
e.Canceled =
true
;
}
}
<
MasterTableView
..............
<CommandItemSettings
AddNewRecordText
=
"Your New text comes here"
/>
Thanks,
Jayesh Goyani
0

Shinu
Top achievements
Rank 2
answered on 10 Oct 2012, 05:47 AM
Hi,
One suggestion is that you can access the buttons and set the PostBackUrl property.
C#:
Thanks,
Shinu.
One suggestion is that you can access the buttons and set the PostBackUrl property.
C#:
protected
void
RadGrid2_ItemDataBound(
object
sender, GridItemEventArgs e)
{
if
(e.Item
is
GridCommandItem)
{
GridCommandItem item = (GridCommandItem)e.Item;
Button btn = (Button)item.FindControl(
"AddNewRecordButton"
);
LinkButton Addbtn = (LinkButton)item.FindControl(
"InitInsertButton"
);
Addbtn.PostBackUrl =
"Page.aspx"
;
btn.PostBackUrl =
"Page.aspx"
;
}
}
Thanks,
Shinu.
0

Robert
Top achievements
Rank 1
answered on 10 Oct 2012, 08:32 AM
Exactly what i was looking for. Thank you very much